# Thread: Question on Fraction with exponent (how it works)

1. ## Question on Fraction with exponent (how it works)

I was going through my math made easy tapes for calculus as I review this(its been 13 years since calc I, and even longer for trig, and alg.)

The professor sets up a formula

g(x) =X^2 + (1)/(x)^2

Now I am not worried about the calc problem I understand that its simple enough.

What confused me is the rule for how he changed 1 divided by X squared to
X raised to the negative 2 power.

(1)/(x)^2 = (X)^-2 could someone break down the relationship as I do not remember this one..

Also perhaps the name of what this is called so I can try to find it on google.

Thanks,

Brian

2. Originally Posted by wilder7bc
I was going through my math made easy tapes for calculus as I review this(its been 13 years since calc I, and even longer for trig, and alg.)

The professor sets up a formula

g(x) =X^2 + (1)/(x)^2

Now I am not worried about the calc problem I understand that its simple enough.

What confused me is the rule for how he changed 1 divided by X squared to
X raised to the negative 2 power. This is from the laws of exponents several semesters ago. I suggest brushing up.

(1)/(x)^2 = (X)^-2 could someone break down the relationship as I do not remember this one..

Also perhaps the name of what this is called so I can try to find it on google.

Thanks,

Brian
.

3. ## Thank you!

That was exactly what I needed... With the name I was able to find a pdf with all the rules

So many things I have forgotten anyway this was:

#7: Negative Law of Exponents: If the base is powered by the
negative exponent, then the base becomes reciprocal with the
positive exponent.

Now I can move forward again, After you get an idea of the boundaries you can then see the relationship between the roots and powers which allows me to prove the law =)

as always this site is great I am going to have to make a donation I think as this is probably one of the best sites I have been on and people are always very helpful.

Thanks again!